iv. cross an albino female, whose father was gray, with a gray male, whose mother was albino.\n1. what is…

iv. cross an albino female, whose father was gray, with a gray male, whose mother was albino.\n1. what is the probability of getting gray offspring?\n2. what is the probability of getting albino offspring?\n3. how many possible genotypes are there among the offspring?\n4. how many possible phenotypes are there among the offspring?\n5. what is the probability of getting heterozygous offspring?\n6. what is the probability of getting homozygous offspring?\n7. what was the genotype of the father of the albino female?

iv. cross an albino female, whose father was gray, with a gray male, whose mother was albino.\n1. what is the probability of getting gray offspring?\n2. what is the probability of getting albino offspring?\n3. how many possible genotypes are there among the offspring?\n4. how many possible phenotypes are there among the offspring?\n5. what is the probability of getting heterozygous offspring?\n6. what is the probability of getting homozygous offspring?\n7. what was the genotype of the father of the albino female?

Answer

Explanation:

Step1: Determine parental genotypes

Let the allele for gray be (A) and for albino be (a). The albino female has genotype (aa) since albino is a recessive trait. Her father was gray, so he must have been (Aa) (to pass on the (a) allele). The gray - male has a mother who was albino ((aa)), so his genotype must be (Aa).

Step2: Set up the Punnett - square

The cross is (Aa\times aa). The possible gametes from (Aa) are (A) and (a), and from (aa) are (a) and (a). The Punnett - square has two rows ((A) and (a) from the male) and two columns ((a) and (a) from the female).

Step3: Calculate probabilities and genotypes

  • The genotypes of the offspring are (Aa) and (aa) in a 1:1 ratio.
  • Probability of gray offspring ((Aa)): The proportion of (Aa) genotypes is (\frac{1}{2}) or 0.5.
  • Probability of albino offspring ((aa)): The proportion of (aa) genotypes is (\frac{1}{2}) or 0.5.
  • Number of possible genotypes: There are 2 possible genotypes ((Aa) and (aa)).
  • Number of possible phenotypes: There are 2 possible phenotypes (gray and albino).
  • Probability of heterozygous offspring ((Aa)): The proportion of (Aa) genotypes is (\frac{1}{2}) or 0.5.
  • Probability of homozygous offspring ((aa)): The proportion of (aa) genotypes is (\frac{1}{2}) or 0.5.
  • Genotype of the father of the albino female: Since the albino female ((aa)) received one (a) allele from each parent and her father was gray, his genotype was (Aa).

Answer:

  1. 0.5
  2. 0.5
  3. 2
  4. 2
  5. 0.5
  6. 0.5
  7. (Aa)